Read this. If the sequence is short exact, then it already has arrows $0 \to \dots$ and $\dots \to 0$ which imply injectivity and surjectivity respectively, right? So in that case there is already always such a map $t$. So what am I getting wrong?
2026-03-29 12:31:44.1774787504
Understanding splitting lemma
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Write $$ 0 \to A \xrightarrow{q} B \xrightarrow{r} C \to 0. $$
The first and last arrows imply injectivity and surjectivity of $q$ and $r$, respectively. This does not imply that there is a splitting.
Mathworld gives a counterexample: $$ 0 \to \Bbb{Z}_2 \to \Bbb{Z}_4 \to \Bbb{Z}_2 \to 0 $$ where the first map takes $1\mapsto 2$. This is not a split SES because $\Bbb{Z}_4$ is not isomorphic to $\Bbb{Z}_2\oplus\Bbb{Z}_2$.
My geometric intuition is better than my algebraic intuition, so I tend to think of short exact sequences as analogues of fiber bundles. A split SES is special: It is like a trivializable fiber bundle. In general, one should not expect a SES to split; $B$ is generically formed by "twisting" $A$ and $C$ together.